#include <stdio.h>

/*If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

Find the sum of all the multiples of 3 or 5 below 1000.*/
int main(int argc, char **argv) {
	int limit = atoi (argv[1]);
	int sum = 0;
	printf("%d\n", limit);
	int i;
	for (i = 0; i < limit; i++) {
		if (i % 3 == 0 || i % 5 == 0) {
			sum += i;
		}
	}
	printf("%d\n", sum);
}
